Consider the function `f(x) = In(sqrt(1-x^2)-x)` then which of the following is/are true?

Options

A. f(x) increases in the on `x=(-1,-(1)/(sqrt2))`
B. f has local maximum at `x=-(1)/(sqrt2)`
C. Least value of f does not exist
D. Least value of f exists

Explanation

To solve the problem, we will analyze the function \( f(x) = \ln(\sqrt{1 - x^2} - x) \) and determine its properties, including its domain, monotonicity, and extrema.

### Step 1: Determine the Domain of \( f(x) \)

The function \( f(x) \) is defined when the argument of the logarithm is positive, i.e.,

\[ \sqrt{1 - x^2} - x > 0 \]

This can be rearranged to:

\[ \sqrt{1 - x^2} > x \]

Squaring both sides (valid since both sides are non-negative for \( x \in (-1, 1) \)) gives:

\[ 1 - x^2 > x^2 \]

This simplifies to:

\[ 1 > 2x^2 \quad \Rightarrow \quad x^2 < \frac{1}{2} \quad \Rightarrow \quad -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \]

Thus, the domain of \( f(x) \) is:

\[ x \in \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \]

### Step 2: Find the Derivative \( f'(x) \)

To analyze the monotonicity, we need to find the derivative \( f'(x) \):

Using the chain rule and quotient rule, we have:

\[ f'(x) = \frac{d}{dx} \ln(\sqrt{1 - x^2} - x) = \frac{1}{\sqrt{1 - x^2} - x} \cdot \left( \frac{-x}{\sqrt{1 - x^2}} - 1 \right) \]

This simplifies to:

\[ f'(x) = \frac{-x - (\sqrt{1 - x^2} - x)}{\sqrt{1 - x^2} - x} = \frac{-\sqrt{1 - x^2}}{\sqrt{1 - x^2} - x} \]

### Step 3: Analyze the Sign of \( f'(x) \)

The sign of \( f'(x) \) depends on the numerator and the denominator:

- The numerator \( -\sqrt{1 - x^2} < 0 \) for \( x \in (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) \).
- The denominator \( \sqrt{1 - x^2} - x \) is positive for \( x < \frac{1}{\sqrt{2}} \).

Thus, \( f'(x) < 0 \) for \( x \in \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \), indicating that \( f(x) \) is decreasing in this interval.

### Step 4: Find Critical Points and Extrema

To find critical points, we set \( f'(x) = 0 \):

\[ -\sqrt{1 - x^2} = 0 \quad \Rightarrow \quad x = \pm 1 \]

However, these points are outside the domain of \( f(x) \). Therefore, we check the endpoints of the domain:

- As \( x \to -\frac{1}{\sqrt{2}} \), \( f(x) \) approaches a finite value.
- As \( x \to \frac{1}{\sqrt{2}} \), \( f(x) \to -\infty \).

Since \( f(x) \) is decreasing throughout its domain and has no local minima, we conclude that:

- \( f(x) \) has a local maximum at \( x = -\frac{1}{\sqrt{2}} \).
- There are no local minima.

### Conclusion

The function \( f(x) \) is defined on the interval \( \left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right) \), is decreasing throughout this interval, has a local maximum at \( x = -\frac{1}{\sqrt{2}} \), and does not have any local minima.